What is the Magnitude of the Magnetic Field for a Proton Beam?

AI Thread Summary
The discussion focuses on calculating the magnitude of the magnetic field for a proton beam traveling at 1.20 km/s, perpendicular to the field. Participants explore the relationship between force, charge, velocity, and magnetic field using relevant equations, including F = qv × B and centripetal force concepts. They discuss deriving the radius of the proton's path from the arc length formula, leading to the calculation of acceleration and ultimately the magnetic field strength. The final correct value for the magnetic field is determined to be approximately 1.5115 mT. The conversation emphasizes the importance of understanding the physics principles involved in the calculations.
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Homework Statement



A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction. The beam travels a distance of 1.30 cm while in the field.

YF-27-24.jpg


What is the magnitude of the magnetic field?

Homework Equations



F = qv ×B
dF = Idl ×B


The Attempt at a Solution



F = qv x B
B = F/qv
= F / (1.602 × 10−19) 1.2

stuck as to what equation applies, and where i can find the value of F from.
 
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the second equation deals with forces on current carrying wires...so you would deal with the first. F is a force...what else in physics is F (Force) equal to...also think of the motion of the proton...it follows a _________ path use that to equate force.
 
ok so:

F = qv x B
B = F/qv
= F / (1.602 × 10^−19) x 1.2 (charge of a proton x velocity of proton)

F = ma?
= 1.67x10^-27 x a (mass of a proton x acceleration)

however, as:
m(v/t) / q(v)
= m(1/t) / q
= (1.67x10^-27 x (1/t)) / 1.602 × 10^−19 (where t = distance / velocity??)
 
i got 1.129x10^-13 and failed my first attempt.
 
how can you relate acceleration with what you are given in the problem we are given "v" and we are also given "r". the acceleration is purely centripetal so the velocity does not change
 
the only other thing is it doesn't say it made a radius of 1.3 so you would have to find it using arc length formula

I hope this is helping you
 
yeh i was about to ask how you knew the radius. ok i'll work on those and get back to you in a few mintues.

**digs deep into the memory bank for equations**
 
would i be correct in saying that using the arc length formula:

pi/2 radians is equivalent to 1.3cm
θ = s / r
r = s / θ
= 1.3 / pi/2
r = 0.8276 cm

t = 0.013m / 1200m/s
= 1.083x10^-5 s

then w = θ / t
= (pi/2) / 1.083x10^-5
= 145041.2121

w = w0 + at
145041.2121 = 0 + a x 1.083x10^-5
a = 1.339x10^10 ??

therefore:

B = F/qv
= ma / qv
= (1.67x10^-27 x 1.339x10^10) / ((1.602×10^−19) x 1200)
= 0.1163 T

or using w instead of v, 9.6237x10^-4put both in and got it wrong again sigh, where am i going wrong?
 
  • #10
r = mv/qb

0.008276 = (1.67x10^-27 x 1200) / (1.602×10^−19 x b)

(1.602×10^−19 x b) = (1.67x10^-27 x 1200) / 0.008276)

(1.602×10^−19 x b) = 2.42146x10^-22

b = 1.5115x10^-3
 
  • #11
that was correct :)

thanks heaps for your help
 
  • #12
Here the force is centripetal force.
SO qvB= mv^2/r
B = mv/qr
substitute the values. See how much you get.
 
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